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Chapter 19: Problem 35

Equipotential surface \(A\) has a potential of \(5650 \mathrm{~V},\) while equipotential surface \(B\) has a potential of \(7850 \mathrm{~V}\). A particle has a mass of \(5.00 \times 10^{-2} \mathrm{~kg}\) and a charge of \(+4.00 \times 10^{-5} \mathrm{C} .\) The particle has a speed of \(2.00 \mathrm{~m} / \mathrm{s}\) on surface \(A .\) An outside force is applied to the particle, and it moves to surface \(B\), arriving there with a speed of \(3.00 \mathrm{~m} / \mathrm{s}\) How much work is done by the outside force in moving the particle from \(A\) to \(B ?\)

Short Answer

Expert verified
The work done by the external force is 0.213 J.

Step by step solution

01

Identify the change in potential energy

The change in the potential energy (abla PE) of the particle is calculated using the particle's charge and the difference in potentials between surfaces A and B. The formula for potential energy change is: \[ \Delta PE = q \times (V_B - V_A)\]Where:- \(q = +4.00 \times 10^{-5} \mathrm{~C}\) is the charge of the particle- \(V_B = 7850 \mathrm{~V}\) is the potential at surface B- \(V_A = 5650 \mathrm{~V}\) is the potential at surface APlugging in the values:\[ \Delta PE = 4.00 \times 10^{-5} \times (7850 - 5650) = 0.088 \mathrm{~J}\]
02

Determine the change in kinetic energy

The kinetic energy (KE) of a particle is given by the formula:\[ KE = \frac{1}{2} m v^2\]Calculate the change in kinetic energy from surface A to B:- Initial speed on A: \(v_A = 2.00 \mathrm{~m/s}\)- Final speed on B: \(v_B = 3.00 \mathrm{~m/s}\)- Mass of the particle: \(m = 5.00 \times 10^{-2} \mathrm{~kg}\)Initial kinetic energy at A:\[ KE_A = \frac{1}{2} \times 5.00 \times 10^{-2} \times (2.00)^2 = 0.10 \mathrm{~J}\]Final kinetic energy at B:\[ KE_B = \frac{1}{2} \times 5.00 \times 10^{-2} \times (3.00)^2 = 0.225 \mathrm{~J}\]Change in kinetic energy:\[ \Delta KE = KE_B - KE_A = 0.225 - 0.10 = 0.125 \mathrm{~J}\]
03

Calculate the total work done

The work done by the external force is equal to the change in total mechanical energy, which is the sum of changes in potential energy and kinetic energy:\[ W_{external} = \Delta KE + \Delta PE\]Plugging in the values from Step 1 and Step 2:\[ W_{external} = 0.125 + 0.088 = 0.213 \mathrm{~J}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equipotential Surface
An equipotential surface is a concept in physics where every point on the surface has the same electric potential. This means that no work is required to move a charge along the surface, as the potential energy does not change. In this exercise, surface A has a potential of 5650 V, while surface B is at 7850 V. The difference in potential between these surfaces indicates that an external force is needed to move a charged particle from one to the other.

It's crucial to remember that the electric force acting on the particle is perpendicular to the equipotential surfaces. Therefore, if a particle moves along an equipotential surface, it does not experience a change in potential energy due to electric forces alone. This movement, however, will change as it moves to a surface with a different potential, requiring external force or other energy changes to facilitate this transition.
Change in Kinetic Energy
Kinetic energy describes the energy that an object possesses due to its motion. For any particle with mass, its kinetic energy is given by the formula: \[ KE = \frac{1}{2} m v^2 \]Where \( m \) is the mass and \( v \) is the velocity of the particle. In this exercise, the particle starts with an initial speed of 2.00 m/s at surface A and reaches a final speed of 3.00 m/s at surface B.

The change in kinetic energy is calculated based on these velocity changes:- Initial Kinetic Energy at A: 0.10 J- Final Kinetic Energy at B: 0.225 J

Thus, the change in kinetic energy (\( \Delta KE \)) is 0.125 J. This increase reflects the work done by the external force to speed up the particle as it moves between the two equipotential surfaces.
Change in Potential Energy
Potential energy is related to the position of a particle within a force field. In an electrostatic context, it is determined by both the charge of the particle and the electric potential it is subject to. The change in potential energy when moving from surface A (5650 V) to surface B (7850 V) is calculated using:\[ \Delta PE = q \times (V_B - V_A) \]Where \( q \) is the charge of the particle.

Given:- Charge \( q = +4.00 \times 10^{-5} \text{ C} \)- Initial potential \( V_A = 5650 \text{ V} \)- Final potential \( V_B = 7850 \text{ V} \)
The change results in \( \Delta PE = 0.088 \text{ J} \).

The work done by the external force needs to overcome this change in potential energy, plus any additional work to increase the particle鈥檚 speed鈥攊n this way facilitating the overall transition between the surfaces.

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Most popular questions from this chapter

The potential difference between the plates of a capacitor is \(175 \mathrm{~V}\). Midway between the plates, a proton and an electron are released. The electron is released from rest. The proton is projected perpendicularly toward the negative plate with an initial speed. The proton strikes the negative plate at the same instant that the electron strikes the positive plate. Ignore the attraction between the two particles, and find the initial speed of the proton.

A charge of \(-3.00 \mu \mathrm{C}\) is fixed in place. From a horizontal distance of \(0.0450 \mathrm{~m}, \mathrm{a}\) particle of mass \(7.20 \times 10^{-3} \mathrm{~kg}\) and charge \(-8.00 \mu \mathrm{C}\) is fired with an initial speed of \(65.0 \mathrm{~m} / \mathrm{s}\) directly toward the fixed charge. How far does the particle travel before its speed is zero?

A positive charge of \(+q_{1}\) is located \(3.00 \mathrm{~m}\) to the left of a negative charge \(-q_{2}\). The charges have different magnitudes. On the line through the charges, the net electric field is zero at a spot \(1.00 \mathrm{~m}\) to the right of the negative charge. On this line there are also two spots where the potential is zero. Locate these two spots relative to the negative charge.

A spark plug in an automobile engine consists of two metal conductors that are separated by a distance of \(0.75 \mathrm{~mm}\). When an electric spark jumps between them, the magnitude of the electric field is \(4.7 \times 10^{7} \mathrm{~V} / \mathrm{m}\). What is the magnitude of the potential difference \(\Delta V\) between the conductors?

Concept Questions An electron and a proton, starting from rest, are accelerated through an electric potential difference of the same magnitude. In the process, the electron acquires a speed \(v_{\mathrm{e}},\) while the proton acquires a speed \(v_{\mathrm{p}}\). (a) As each particle accelerates from rest, it gains kinetic energy. Does it gain or lose electric potential energy? (b) Does the electron gain more, less, or the same amount of kinetic energy as the proton does? (c) Is \(v_{\mathrm{e}}\) greater than, less than, or equal to \(v_{\mathrm{p}}\) ? Justify your answers. Find the ratio \(v_{\mathrm{e}} / v_{\mathrm{p}}\). Verify that your answer is consistent with your answers to the Concept Questions.
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